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3 years ago

6

What does an Astronomer study? *

1 answer:

ratelena [41]3 years ago

3 0

Answer:

an astronomer studies planets, stars, moons, etc, or objects that are outside the field of Earth.

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A car with a mass of 1380 Kg is traveling at 23 m/s to the north. A truck with a mass of 1625 Kg is traveling at 26 m/s to the s

trasher [3.6K]

Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum $p_{i}$ must be equal to the final momentum $p_{f}$ , and taking into account this is aninelastic collision:

Before the collision:

$p_{i}=mV_{o}+MU_{o}$ (1)

After the collision:

$p_{f}=(m+M)V_{f}$ (2)

Where:

$m=1380 kg$ is the mass of the car

$V_{o}=23 m/s$ is the velocity of the car, directed to the north

$M=1625 kg$ is the mass of the truck

$U_{o}=-26 m/s$ is the velocity of the truck, directed to the south

$V_{f}$ is the final velocity of both the car and the truck

$p_{i}=p_{f}$ (3)

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1380 kg)(23 m/s)+(1625 kg)(-26 m/s)}{1380 kg+1625 kg}$ (6)

Finally:

$V_{f}=-3.49 m/s$ The negative sign indicates the direction of the velocity is to the south

8 0

3 years ago

Why are simple everyday actions considered

Oxana [17]

They create energy ……………….

8 0

3 years ago

Read 2 more answers

Electrically inert metal ball A is connected to the ground by a wire. What happens to the charge of this ball if you bring a neg

kaheart [24]

Explanation:

They will repel, meaning that they are made of an electrical conductor.

7 0

3 years ago

10.A car is travelling at a constant speed of 27m/s. The driver looks away from the road for a 2.0s to tune in a station on the

Korvikt [17]

Explanation:

Distance = speed × time

d = (27 m/s) (2.0 s)

d = 54 m

5 0

3 years ago

A very long uniform line of charge has charge per unit length λ1 = 4.68 μC/m and lies along the x-axis. A second long uniform li

Kitty [74]

Answer:

$E_{net} = 6.44 \times 10^5 N/C$

Explanation:

As we know that electric field due to infinite line charge distribution at some distance from it is given as

$E = \frac{2k \lambda}{r}$

now we need to find the electric field at mid point of two wires

So here we need to add the field due to two wires as they are oppositely charged

Now we will have

$E_{net} = \frac{2k\lambda_1}{r} + \frac{2k\lambda_2}{r}$

now plug in all data

$\lambda_1 = 4.68 \muC/m$

$\lambda_2 = 2.48 \mu C/m$

$r = 0.200 m$

now we have

$E_{net} = \frac{2k}{r}(4.68 + 2.48)$

$E_{net} = \frac{2(9\times 10^9)}{0.200}(7.16 \times 10^{-6})$

$E_{net} = 6.44 \times 10^5 N/C$

8 0

3 years ago